ch4 - 4.5.IDENTIFY: Vector addition. SET UP: Use a...

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4.5.IDENTIFY: Vector addition. SET UP: Use a coordinate system where the -axis x + is in the direction of , A F G the force applied by dog A. The forces are sketched in Figure 4.5. EXECUTE: 270 N, Ax F =+ 0 Ay F = cos60.0 (300 N)cos60.0 150 N Bx B FF = ° = + sin60.0 (300 N)sin60.0 260 N By B = ° = + Figure 4.5a AB R GG G 270 N 150 N 420 N xA xB x RF F =+= + + = + 0 260 N 260 N yA yB y + = + 22 x y R RR (420 N) (260 N) 494 N R = tan 0.619 y x R R θ == 31.8 = ° Figure 4.5b E VALUATE : The forces must be added as vectors. The magnitude of the resultant force is less than the sum of the magnitudes of the two forces and depends on the angle between the two forces. 4.9.IDENTIFY: Apply m = F a G G to the box. SET UP: Let x + be the direction of the force and acceleration. 48.0 N x F = . EXECUTE: x x Fm a = gives 2 48.0 N 16.0 kg 3.00 m/s x x F m a = . E VALUATE : The vertical forces sum to zero and there is no motion in that direction. 4.11.IDENTIFY and SET UP: Use Newton’s second law in component form (Eq.4.8) to calculate the acceleration produced by the force. Use constant acceleration equations to calculate the effect of the acceleration on the motion. EXECUTE: (a) During this time interval the acceleration is constant and equal to 2 0.250 N 1.562 m/s 0.160 kg x x F a m = We can use the constant acceleration kinematic equations from Chapter 2. 11 00 0 (1.562 m/s )(2.00 s) , xx xx v t a t −= + 2 so the puck is at 3.12 m. x = 2 0 0 (1.562 m/s )(2.00 s) 3.13 m/s. x vv a t + = (b) In the time interval from to 5.00 s the force has been removed so the acceleration is zero. The speed stays constant at The distance the puck travels is At the end of the interval it is at 2.00 s t = 3.12 m/s. x v = (3.12 m/s)(5.00 s 2.00 s) 9.36 m. x t −= = = 0 9.36 m 12.5 m. =
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In the time interval from to 7.00 s the acceleration is again At the start of this interval and 5.00 s t = 2 1.562 m/s . x a = 0 3.12 m/s x v = 0 12.5 m. x = 22 11 00 (3.12 m/s)(2.00 s) (1.562 m/s )(2.00 s) . xx xx v t a t −= + = + 2 0 6.24 m 3.12 m 9.36 m. −= + = Therefore, at the puck is at 7.00 s t = 0 9.36 m 12.5 m 9.36 m 21.9 m. =+ = + = 2 0 3.12 m/s (1.562 m/s )(2.00 s) 6.24 m/s x vv a t =+= + = E VALUATE : The acceleration says the puck gains 1.56 m/s of velocity for every second the force acts. The force acts a total of 4.00 s so the final velocity is (1.56 m/s)(4.0 s) 6.24 m/s. = 4.22. IDENTIFY: m = F a G G refers to forces that all act on one object. The third law refers to forces that a pair of objects exert on each other. SET UP: An object is in equilibrium if the vector sum of all the forces on it is zero. A third law pair of forces have the same magnitude regardless of the motion of either object. EXECUTE: (a) the earth (gravity) (b) 4 N; the book (c) no, these two forces are exerted on the same object (d) 4 N; the earth; the book; upward (e) 4 N, the hand; the book; downward (f) second (The two forces are exerted on the same object and this object has zero acceleration.) (g) third (The forces are between a pair of objects.)
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ch4 - 4.5.IDENTIFY: Vector addition. SET UP: Use a...

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